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需列計算過程，否則不予計分
填充計算題（總計 $10$ 題，每題 $10$ 分）
- 試求級數 $\displaystyle x-\frac{x^2}{2^2}+\frac{x^3}{3^2}-\frac{x^4}{4^2}+\cdots$ 之收斂區間 $=$ 。
- 當 $x=-1$ 時級數寫為 $\displaystyle\sum_{n=1}^{\infty}\frac{-1}{n^2}$，由 $p$ 級數在 $p=2$ 時收斂可知此級數收斂；
- 當 $x=1$ 時級數寫為 $\displaystyle\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n^2}$ 可由交錯級數審歛法知其收斂。
- 星形線(Asteroid) $x=a\cos^3t$, $y=a\sin^3t$ 所圍成之面積 $=$ 。
- 求下列各線所圍成區域之面積 。
$\displaystyle xy=\frac{a^2}2,\quad xy=2a^2,\quad y=\frac{x}2,\quad y=2x$
- 求出此心臟線 $r=1+\sin\theta$ 的長度 $=$ 。
- $\displaystyle\int_0^1\int_{2x}^2e^{y^2}dydx=$ .
- $\displaystyle\int_1^2\frac{e^{1/x}}{x^2}dx=$ .
- If $y=x^{\sqrt{x}}$, then $\displaystyle\frac{dy}{dx}=$ .
- Let $R=\ln\left(u^2+v^2+w^2\right)$, $u=x+2y$, $v=2x-y$, $w=2xy$. When $x=y=1$, $\displaystyle\frac{\partial R}{\partial x}=$ , and $\displaystyle\frac{\partial R}{\partial y}=$ .
- Suppose that $f\left(x\right)$ is differentiable everywhere and that $f\left(0\right)=-3$ and $f'\left(x\right)\leq5$. The largest possible value for $f\left(2\right)$ is .
- The half-life of radium-226 is 1590 years.
- A sample of radium-226 has a mass of $100$ mg. The formula for the mass of the sample that remains after $t$ years, $m\left(t\right)$ is .
- The time taken for the mass of the sample to be reduced to $30$ mg is .
- 按照半衰期的意義可列出對應的質量函數為
$\displaystyle m\left(t\right)=100\cdot\left(\frac12\right)^{\frac{t}{1590}}~(\mbox{mg})$
- 設樣本要減少至 $30$ 毫克所需時間 $t_0$，則按前一小題應滿足
$\displaystyle100\cdot\left(\frac12\right)^{\frac{t_0}{1590}}=30$
因此有 $2^{-\frac{t_0}{1590}}=0.3$，取以 $2$ 為底的對數有$\displaystyle\frac{t_0}{1590}=\log_2\frac{10}3=\log_210-\log_23=\frac{1-\log_{10}3}{\log_{10}2}$
因此所求為$\displaystyle t_0=1590\cdot\frac{1-\log_{10}3}{\log_{10}2}=1590\cdot\frac{1-0.4771}{0.301}=1590\cdot\frac{5229}{3010}\approx2762.1627907$
故約莫 $2762$ 年才能變為 $30$ 毫克。
訣竅
運用比值審歛法的概念求出收斂半徑，隨後再判定其端點的歛散性即可。解法
給定的級數可表示為 $\displaystyle\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n^2}x^n$，如此可知收斂半徑為$\displaystyle R=\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|=\lim_{n\to\infty}\frac{\left(n+1\right)^2}{n^2}=1$
而當訣竅
由參數下的面積公式求解，其中可利用對稱性來簡化計算過程。此題同九十六年應用微積分第七題。解法
所求面積可表達並計算如下$\displaystyle\begin{aligned}A&=\int_{x=0}^{x=a}y\left(t\right)dx\left(t\right)=4\int_{\frac\pi2}^0a\sin^3tdt\cdot-3a\cos^2t\sin tdt=12a^2\int_0^{\frac\pi2}\sin^4t\cos^2tdt\\&=12a^2\int_0^{\frac\pi2}\left(\frac{1-\cos2t}2\right)^2\left(\frac{1+\cos2t}2\right)dt=\frac{3a^2}2\int_0^{\frac\pi2}\left(1-\cos2t\right)\left(1-\cos^22t\right)dt\\&=\frac{3a^2}4\int_0^{\frac\pi2}\left(1-\cos2t\right)\left(1-\cos4t\right)dt=\frac{3a^2}4\int_0^{\frac\pi2}\left(1-\cos2t-\cos4t+\cos2t\cos4t\right)dt\\&=\frac{3a^2}4\cdot\frac\pi2-\frac{3a^2}4\int_0^{\frac\pi2}\cos2t\cos4tdt=\frac{3\pi a^2}8-\frac{3a^2}8\int_0^{\frac\pi2}\left(\cos6t+\cos2t\right)dt\\&=\frac{3\pi a^2}8\end{aligned}$
訣竅
面積可用一個單純的重積分來表示，隨後運用變數代換來處理重積分。解法
設 $D$ 為四曲線所圍成之區域，那麼所求之面積為 $\displaystyle\iint_DdA$。令 $\left\{\begin{aligned}&u=xy\\&v=y/x\end{aligned}\right.$，那麼由邊界條件可知變數範圍為 $\left\{\begin{aligned}&\frac{a^2}2\leq u\leq2a^2\\&\frac12\leq v\leq2\end{aligned}\right.$。再者，其 Jacobian 行列式為$\displaystyle\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}=\left|\begin{matrix}\displaystyle\frac{\partial x}{\partial u}&\displaystyle\frac{\partial x}{\partial v}\\\displaystyle\frac{\partial y}{\partial u}&\displaystyle\frac{\partial y}{\partial v}\end{matrix}\right|=\left|\begin{matrix}\displaystyle\frac{\partial u}{\partial x}&\displaystyle\frac{\partial u}{\partial y}\\\displaystyle\frac{\partial v}{\partial x}&\displaystyle\frac{\partial v}{\partial y}\end{matrix}\right|^{-1}=\left|\begin{matrix}y&x\\\displaystyle-\frac{y}{x^2}&\displaystyle\frac1x\end{matrix}\right|^{-1}=\frac{2y}x=2v$
如此所求的面積為$\displaystyle A=\iint_DdA=\int_{\frac{a^2}2}^{2a^2}\int_{\frac12}^22vdvdu=\frac{3a^2}2\cdot\frac{15}4=\frac{45a^2}8$
訣竅
運用極座標曲線下的長度公式與對稱性求解即可。解法
運用極座標下的曲線長度公式與對稱性可以計算如下$\displaystyle s=\int_{-\pi}^{\pi}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta=\int_{-\pi}^{\pi}\sqrt{\left(1+\sin\theta\right)^2+\left(\cos\theta\right)^2}d\theta=\int_{-\pi}^{\pi}\sqrt{2+2\sin\theta}d\theta=2\int_{-\frac\pi2}^{\frac\pi2}\sqrt{2+2\sin\theta}d\theta$
令 $\displaystyle\psi=\theta+\frac\pi2$，那麼所求的曲線弧長可改寫並計算如下$\displaystyle s=2\int_0^{\pi}\sqrt{2+2\sin\left(\psi-\frac\pi2\right)}d\psi=2\int_0^{\pi}\sqrt{2-2\cos\psi}d\psi=4\int_0^{\pi}\sin\frac{\psi}2d\psi=\left.-8\cos\frac\psi2\right|_0^{\pi}=8$
訣竅
交換積分次序後求解。此題同九十五年應用微積分第五題。解法
原積分範圍 $\left\{\begin{aligned}&0\leq x\leq1\\&2x\leq y\leq2\end{aligned}\right.$ 可改寫為 $\left\{\begin{aligned}&0\leq x\leq y/2\\&0\leq y\leq2\end{aligned}\right.$，如此所求的重積分可改寫並計算如下$\displaystyle\int_0^1\int_{2x}^2e^{y^2}dydx=\int_0^2\int_0^{y/2}e^{y^2}dxdy=\int_0^2\frac{ye^{y^2}}2dy=\left.\frac{e^{y^2}}4\right|_0^2=\frac{e^4-1}4$
訣竅
運用變數代換的思想直接計算即可。解法
直接計算可知$\displaystyle\int_1^2\frac{e^{1/x}}{x^2}dx=-\int_1^2e^{1/x}d\left(\frac1x\right)=-e^{\frac1x}\Big|_1^2=e-e^{\frac12}$
訣竅
換底後使用連鎖律求導即可。解法
換底後使用連鎖律求導如下$\displaystyle\frac{dy}{dx}=\frac{d}{dx}x^{\sqrt{x}}=\frac{d}{dx}e^{\sqrt{x}\ln x}=e^{\sqrt{x}\ln x}\frac{d}{dx}\left(\sqrt{x}\ln x\right)=x^{\sqrt{x}}\left(\frac{\ln x}{2\sqrt{x}}+\frac1{\sqrt{x}}\right)=\frac{x^{\sqrt{x}}\left(2+\ln x\right)}{2\sqrt{x}}$
訣竅
使用多變函數連鎖律即可。解法
使用多變數函數的連鎖律可知$\displaystyle\begin{aligned}&\frac{\partial R}{\partial x}=\frac{\partial R}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial R}{\partial v}\frac{\partial v}{\partial x}+\frac{\partial R}{\partial w}\frac{\partial w}{\partial x}=\frac{2u}{u^2+v^2+w^2}\cdot1+\frac{2v}{u^2+v^2+w^2}\cdot2+\frac{2w}{u^2+v^2+w^2}\cdot2y=\frac{2u+4v+4wy}{u^2+v^2+w^2}\\&\frac{\partial R}{\partial y}=\frac{\partial R}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial R}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial R}{\partial w}\frac{\partial w}{\partial y}=\frac{2u}{u^2+v^2+w^2}\cdot2+\frac{2v}{u^2+v^2+w^2}\cdot\left(-1\right)+\frac{2w}{u^2+v^2+w^2}\cdot2x=\frac{4u-2v+4wx}{u^2+v^2+w^2}\end{aligned}$
而當 $x=y=1$ 時有 $u=3$、$v=1$、$w=2$，因此$\displaystyle\frac{\partial R}{\partial x}\left(1,1\right)=\frac{2\cdot3+4\cdot1+4\cdot2\cdot1}{3^2+1^2+2^2}=\frac97,\qquad\frac{\partial R}{\partial y}\left(1,1\right)=\frac{4\cdot3-2\cdot1+4\cdot2\cdot1}{3^2+1^2+2^2}=\frac97$
訣竅
運用微積分基本定理估算其最大值。解法
利用微積分基本定理可知$\displaystyle f\left(2\right)=f\left(0\right)+\int_0^2f'\left(s\right)ds\leq-3+\int_0^25ds=-3+10=7$
故 $f\left(2\right)$ 的最大值為 $7$。
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